2,R>0$ such that \begin{equation}\label{cinque} 0<\nu G(x,s)\leq s g(x,s). \end{equation} for almost every $x\in \Omega $ and all $s \in \mathbb{R}$ with $|s|\geq R$. It easily follows that the function $a$ satisfies the following conditions: \begin{itemize} \item there exist $c_1,c_2>0$ such that \begin{equation}\label{uno} \frac{c_1}{{(1+|s|)}^{2\alpha}}\leq a(x,s)\leq c_2 \end{equation} for almost every $x \in \Omega$, for every $s \in \mathbb{R}$, \item for almost every $x$ in $\Omega$ the function $a(x,.)$ is differentiable on $\mathbb{R}$ and there exist $c_3>0$ such that, for almost every $x \in \Omega$, its derivative $$ a_s(x,s)\equiv \frac{\partial a}{\partial s}(x,s)=\frac{-2\alpha\,s\,a(x,s)}{b(x)+s^2} $$ satisfies \begin{equation}\label{due} |a_s(x,s)|\leq c_3 \quad \forall s \in \mathbb{R} \end{equation} \item for almost every $x \in \Omega$ and all $s \in \mathbb{R}$ \begin{equation}\label{quattro1} a(x,s)=a(x,-s). \end{equation} \end{itemize} \begin{definition} \label{def1.1} \rm We say that $u$ is a weak solution of \eqref{eq:q} if $u \in H_0^1(\Omega)$ and $$ \int_\Omega j_\xi(x,u,\nabla u)\nabla v+j_s(x,u,\nabla u)v =\int_\Omega g(x,u)\,v $$ for every $v \in C_0^\infty(\Omega)$. \end{definition} We are now able to state our main result. \begin{theorem}\label{undicio} Assume that conditions \eqref{quattro}, \eqref{alfa} and \eqref{cinque} hold. Then there exists a sequence $\{u_h\}\subset H_0^1(\Omega)\cap L^\infty(\Omega)$ of weak solutions of (\ref{eq:q}) such that $$ \int_\Omega j(x,u_h,\nabla u_h)-\int_\Omega G(x,u_h) $$ approaches $+\infty$ as $h\to +\infty$. \end{theorem} \section{Proof of the main result} Let $\varphi \in C^2(\mathbb{R})$ be defined as $$ \varphi(s)=s\,{(1+s^2)}^{\frac{\alpha}{2(1-\alpha)}}. $$ \begin{remark}\label{nota4} \rm We observe that $\varphi$ is odd and that there exists $\gamma>0$ such that \begin{equation}\label{one} \varphi'(s)\geq \gamma(1+|\varphi(s)|)^\alpha. \end{equation} Moreover we have \begin{equation}\label{ai} \lim_{s\to \pm \infty} \frac{s\,\varphi'(s)}{\varphi(s)}= \lim_{s\to \pm \infty} \Big(1+s\frac{\varphi''(s)}{\varphi'(s)}\Big)=\frac{1}{1-\alpha}. \end{equation} \end{remark} Let us consider the change of variable $u=\varphi(v)$. We can define on $H_0^1(\Omega)$ the functional $$ \tilde{f}(v)=\frac{1}{2}\int_\Omega A(x,v)|\nabla v|^2- \int_\Omega \widetilde{G}(x,v), $$ where $$ A(x,s)=a(x,\varphi(s))\cdot (\varphi'(s))^2, \quad \widetilde{G}(x,s)=G(x,\varphi(s))=\int_0^s \tilde{g}(x,t)dt, $$ with $\tilde{g}(x,s)=g(x,\varphi(s))\cdot \varphi'(s)$. Now let us consider the integrand $\tilde{j}: \Omega\times \mathbb{R}\times {\mathbb{R}}^n\to\mathbb{R}$ defined by $$ \tilde{j}(x,s,\xi)=\frac{1}{2}A(x,s)|\xi|^2. $$ \begin{remark}\label{notaz2} \rm It is readily seen that \eqref{one} and the left inequality of (\ref{uno}) imply that for almost every $x \in \Omega$ and for every $(s,\xi)\in \mathbb{R}\times {\mathbb{R}}^n$, there holds $$ \tilde{j}(x,s,\xi)\geq \alpha_0|\xi|^2, $$ where $\alpha_0=\frac{c_1 \,\gamma^2}{2}$. \end{remark} \begin{remark}\label{notazz} \rm By Remark \ref{nota4} there exists $\Lambda>0$ such that, for a.e. $x\in \Omega$ and for every $s \in \mathbb{R}$, we have \begin{gather}\label{del} A(x,s)\leq \Lambda, \\ \label{del1} |A_s(x,s)|\leq \Lambda. \end{gather} \end{remark} \begin{proposition}\label{mill1} Assume condition \eqref{cinque}. Then \begin{equation}\label{cedo} \frac{\nu}{1-\alpha}\,\widetilde{G}(x,s)\leq s\tilde{g}(x,s) \end{equation} for every $s\in \mathbb{R}$ with $|s|\geq R$. \end{proposition} \begin{proof} Condition (\ref{cinque}) implies $$ \nu \widetilde{G}(x,s)\,\varphi'(s)\leq \varphi(s)\, \widetilde{G}_s(x,s) $$ hence $$ \nu \frac{s \varphi'(s)}{\varphi(s)}\widetilde{G}(x,s)\leq s \widetilde{G}_s(x,s) $$ and taking into account Remark \ref{nota4}, we get the thesis. \end{proof} \begin{proposition}\label{notaz4} There exists $\mu< \frac{\nu}{1-\alpha}-2$ such that, for almost every $x \in \Omega$, for every $\xi \in {\mathbb{R}}^n$, for every $s \in \mathbb{R}$ with $|s|\geq R$, we have \begin{equation}\label{cici} 0\leq s\tilde{j}_s(x,s,\xi)\leq \mu\, \tilde{j}(x,s,\xi). \end{equation} \end{proposition} \begin{proof} Indeed \begin{equation} \label{uu} \begin{aligned} \tilde{j}_s(x,s,\xi) &=\frac{1}{2}A_s(x,s)|\xi|^2\\ &= \frac{1}{2}[a_s(x,\varphi(s))\cdot (\varphi'(s))^3+2\varphi'(s)\cdot\varphi''(s)\cdot a(x,\varphi(s))]|\xi|^2\\ &=a(x,\varphi(s))\cdot \varphi'(s)\Big[\frac{-\alpha \,\varphi(s)\, (\varphi'(s))^2}{b(x)+(\varphi(s))^2}+\varphi''(s)\Big]|\xi|^2. \end{aligned} \end{equation} Let $s>0$. Then recalling that $a(x,\varphi(s))$ and $\varphi'(s)$ are positive functions, it suffices to prove that the square bracket is non negative. Note that the expression is equal to $$ \frac{\alpha s{(1+s^2)}^{\frac{\alpha}{2(1-\alpha)}}}{(1-\alpha){(1+s^2)}^2} \Big[\frac{-(s^2+1-\alpha){(1+s^2)}^{\frac{\alpha}{(1-\alpha)}}}{b(x) +{(1+s^2)}^{\frac{\alpha}{(1-\alpha)}}\,s^2}+ \frac{(s^2+1-\alpha)\,b(x)}{(1-\alpha)(b(x)+{(1+s^2)} ^{\frac{\alpha}{1-\alpha}}\,s^2)}+2\Big]. $$ Observing that the second term in square bracket is positive and the sum of the first and third is equal to $$ \frac{{(1+s^2)}^{\frac{\alpha}{1-\alpha}}(s^2-(1-\alpha)) +2\,b(x)}{b(x)+{(1+s^2)}^{\frac{\alpha}{1-\alpha}}\,s^2}, $$ the assertion follows if we assume $R=\sqrt{1-\alpha}$. On the other hand if $s<0$, taking into account that $\varphi''(s)$ is an odd function, we deduce that the square bracket in (\ref{uu}) is negative. Now we prove the right inequality. Since $\varphi'(s)\geq 0$ and $a_s(x,\varphi(s))\leq 0$, we have $$ s\,\tilde{j}_s(x,s,\xi)\leq 2\,\tilde{j}(x,s,\xi)\Big(\frac{\varphi''(s)\,s}{\varphi'(s)}\Big) $$ and by Remark \ref{nota4} it follows the assertion with $R$ large enough and $\mu= \frac{2\,\alpha}{1-\alpha}$. \end{proof} We now are able to prove the main result of the paper. \begin{proof}[Proof of Theorem \ref{undicio}] By Remark \ref{notaz2} and \ref{notazz} and Proposition \ref{mill1} and \ref{notaz4} we are able to apply Theorem 2.6 in \cite{Ca}. So obtaining a sequence of weak solutions $\{v_h\} \subset H_0^1(\Omega)$ of the problem $$ \int_\Omega A(x,v)\nabla v \nabla w+ \frac{1}{2}\int_\Omega A_s(x,v)|\nabla u|^2\,w= \int_\Omega \tilde{g}(x,v)w $$ for every $w \in C_0^\infty(\Omega)$ with $f(v_h)\to\infty$. By Theorem 7.1 in \cite{Pe-Sq} these solutions belong to $L^\infty(\Omega)$. If we set $u_h=\varphi(v_h)$, it is clear that $u_h\in H_0^1(\Omega)\cap L^\infty(\Omega)$ and an easy calculation shows that each $u_h$ is a weak solution of (\ref{eq:q}) with $$ \int_\Omega j(x,u_h,\nabla u_h)-\int_\Omega G(x,u_h) \to +\infty $$ as $h\to +\infty$. \end{proof} \begin{thebibliography}{0} \bibitem{Ar1} D. Arcoya, L. Boccardo, \emph{Critical points for multiple integrals of the calculus of variations.} Arch. Rational Mech. Anal., \textbf{134}(1996),249--274. \bibitem{Ar2} D.Arcoya, L. Boccardo, \emph{Some remarks on critical point theory for nondifferentiable functionals}, NoDEA Nonlinear Differential Equations Appl. \textbf{6} (1999), 79-100. \bibitem{ar-bo-or} D. Arcoya, L. Boccardo, L. Orsina, \emph{Existence of critical points for some noncoercive functionals}, Ann. I. H. Poincar\'{e} Anal. Nonlin\'{e}aire., \textbf{18},no.4 (2001), 437-457. \bibitem{Ca} A. 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